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# A $10$ henry inductor is carrying current of$2A$ . At what rate should the current be changed so that an emf of $100V$ induced?$\left( A \right)5A{s^{ - 1}}$  $\left( B \right)10A{s^{ - 1}}$ $\left( C \right)15A{s^{ - 1}}$ $\;\left( D \right)20A{s^{ - 1}}$

Last updated date: 21st Jul 2024
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Hint: The relation between inductance and current should be used. Note that, the induced emf of an inductor is proportional to the rate of change of current, where the inductance is the proportional constant.
Put the values of the given entities in the above formula and find the required rate.

Formula used:
The induced emf in the inductor $E = - L\dfrac{{dI}}{{dT}}$
Where,
$L =$ self-inductance
$\dfrac{{dI}}{{dT}} =$ rate of change of current.

Self-inductance is the result of the device's causation electrical phenomenon in itself.
electrical phenomenon The induced emf in the inductor $E = - L\dfrac{{dI}}{{dT}}$
, wherever $L$ is that self-inductance of the inductor,
and $\dfrac{{dI}}{{dT}} =$ the rate of change of current through it.
The sign indicates that the electrical phenomenon opposes the modification in current.
So if we neglect the sign, the formula will be $E = L\dfrac{{dI}}{{dT}}$
Given, $E = 100V$
$L = 10H$
$dI = 2A$
$\therefore 100 = \dfrac{{10 \times 2}}{{dT}}$
$\Rightarrow dT = \dfrac{1}{5}$
So the rate, $\dfrac{{dI}}{{dT}} = \dfrac{2}{{\dfrac{1}{5}}}$
$\Rightarrow \dfrac{{dI}}{{dT}} = 10A{s^{ - 1}}$
So, the rate $10A{s^{ - 1}}$

So, the correct answer is “Option B”.

Note: According to Lenz's law, the induced voltage has a polarity that opposes the change in current through it. As a result, inductors oppose any changes in current through them. An inductor is characterized by its inductance, which is the ratio of the voltage to the rate of change of current.
When this current flowing through the inductance changes, the time-varying flux induces electrical phenomenon (e.m.f.) (voltage) within the conductor, delineated by Faraday's law of induction. ... As a result, inductors oppose any changes in current through them.